For (constr.) KA, which is drawn through the centre K, cuts BC at right angles in A; ...(E. 3.3.) BA=AC. PROP. III. 4. THEOREM. If two isosceles triangles be of equal altitudes, and the side of the one be equal to the side of the other, their bases shall be equal. Let BKC, DKE be two isosceles, having B K E either of the equal sides, as BK, of the one, equal to either of the equal sides, as DK, of the other, and having, also, their altitudes, that is, the perpendiculars drawn the vertex to the base in each, equal to one another: The base BC is equal, also, to the base DE. For the being supposed to be so placed as to have their summits in the same point K, if they wholly coincide, it is manifest, that BC=DE; And, if they do not coincide, since (hyp.) KB = KD=KE= KC, a circle described from the centre K, at the distance KB, will pass through D, E, and C. From K as a centre, describe, therefore, the circle BDEC; and since (hyp.) the KA= 1 KL, .. (E. 14. 3.) BC=DE. PROP. IV. 5. THEOREM. Any two chords of a circle which cut a diameter in the same point and at equal angles, are equal to one another. Let any two chords AB, CD of the circle ADBC with it the cut a diameter. EF in the same point G, and make AGE = CGE: Then AB-CD. K, the centre of the circle, draw For, from (E. 12. 1.) KL to AB, and KM to CD. And since (hyp. and E. 15.1.) the LGK =/ MGK, and (constr.) the at L and M are right angles, and that KG is common to the two KLG, KMG, (E. 26.1.) KL KM; ... (E. 14.3.) AB = CD. 6. PROBLEM. PROP. V. Through a given point, within a given circle, to draw two equal chords, making with one another an angle equal to a given rectilineal angle. Let G be a given point in the circle ADBC, and XHY a given rectilineal angle: It is required to draw through G two equal chords of the circle ADBC, which shall make with one another an <=2XHY. Find the centre K (E. 1. 3.) of the circle ADBC, and join K, G; bisect (E. 9. 1.) the XHY by ZH; at the point G, in KG, make (E. 23. 1.) the KGB, KGD each equal to the ZHX or ZHY, and produce BG and DG to meet the circumference in A and C respectively. Then (constr.) the whole BGD = 4 XHY; and since (constr.) the KGB= ▲ KGD, ... (S. 4. 3.) the chord AB= chord CD. PROP. VI. 7. THEOREM. If the diameters of two circles are in the same straight line, and have a common extremity, the two circles shall touch one another. For since (hyp.) the two diameters are in the same straight line, it is manifest that a straight line drawn, from their common extremity, perpendicular to either of them will be perpendicular to the other, and ... (E. 16. 3. cor.) will touch both the circles: The circles, .., (E. 3. def. 3.) will touch one another: For it is plain that they cannot cut one another without also cutting the straight line that has been shewn to be their common tangent; which is impossible. PROP. VII. 8. PROBLEM. Three points being given in the circumference of a circle, and the middle point being equidistant from the other two, to describe two equal circles; which shall touch each other in the middle point, and which shall pass the one through one of the extreme points, and the other through the other extreme point. Let A, B, C, be three given points in the cir cumference of the circle ABC, and let the middle point, B, be equidistant from A and B: It is required to describe two equal circles, the one passing through A and the other through C, which shall touch one another in B. Join A, B, and B, C; find (E. 1. 3.) the centre K of the circle; from K draw (E. 12. 1.) KD 1 to AB, and KE to BC; join K, B; and through B draw (E. 11. 1.) FBG to KB meeting KD and KE, produced in F and G respectively. Then since (hyp.) AB=BC, .. (E. 14. 3.) KD KE; and KB is common to the two KDB and KEB, and (hyp. constr. and E. 3. 3.) the side DB the side EB; ... (E. 8. 1.) the ▲ DKB= / EKB. L And, since KB is common to the two KBF, KBG, and the FKB=2 EKB, and (constr.) the I |